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$$\begin{bmatrix} 4 & 1 & 1 \\ 1 & 2 & -1 \\ 1 & -1 & 3 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 3 \\ 1 \\ \frac{3}{2} \end{bmatrix}$$

Show that the matrix above is positive definite and solve the system using the decomposition $A = LDL^T$ with $L$ unitary angular and $D$ diagonal

I think I can use the decomposition below:

enter image description here

So we have

$$\begin{bmatrix} 4 & 1 & 1 \\ 1 & 2 & -1 \\ 1 & -1 & 3 \end{bmatrix} = \begin{bmatrix} D_1 \\ L_{21}D_1 & L_{21}^2D_1+D_2 \\ L_{31}D_1 & L_{31}L_{21}D_1+L_{32}D_2 & L_{31}^2D_1+L_{32}^2D_2 + D_3 \end{bmatrix}$$

But what do I do with the other coefficients of the matrix which are not equal to the right side? I mean, what do I do with $a_{12}, a_{13}$ for example?

And how do I proceed to solve the system?

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If you have $LDL^Tx=b$, multiply on the left by $L^T$ to get $$ DL^Tx=L^Tb. $$ So you have a system $Dy=L^Tb$, with $D$ diagonal; so you simply have $y_j=(L^Tb)_j/D_{jj}$ for each $j$. Since $y=L^Tx$, you have $x=Ly$, which you can calculate directly.

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  • $\begingroup$ Why multplying on the left by $L^T$ cancels $L$? $\endgroup$ – Guerlando OCs Apr 4 at 23:11
  • $\begingroup$ Because $L$ is unitary. That's what you wrote? $\endgroup$ – Martin Argerami Apr 5 at 0:11

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